Partial fractions (5x^2+1)/[(3x+2)(x^2+3)]

In summary: A is to be divided by 3x+ 2 so A= -3/2. Bx+ c is to be divided by x^2+ 3 so Bx+ c= 2x+ 2 and B= 2, c= 2. In summary, the partial fraction decomposition of $\frac{5x^2+1}{(3x+2)(x^2+3)}$ is $\frac{-3/2}{3x+2}+\frac{2x+2}{x^2+3}$.
  • #1
mathlearn
331
0
Trouble here in the below partial fraction (Bug)

$\frac{5x^2+1}{(3x+2)(x^2+3)}$

One factor in the denominator is a quadratic expression

Split this into two parts A&B

$\frac{5x^2+1}{(3x+2)(x^2+3)}=\frac{A}{(3x+2)}+\frac{Bx+c}{(x^2+3)}$

$\frac{5x^2+1}{(3x+2)(x^2+3)}=\frac{A(x^2+3)}{(3x+2)}+\frac{Bx+c (3x+2)}{(x^2+3)}$

${5x^2+1}={A(x^2+3)}+{Bx+c (3x+2)}{}$

and cannot take it from here
 
Mathematics news on Phys.org
  • #2
Hi mathlearn,

I believe you have a typo in your last line, it should be:
${5x^2+1}={A(x^2+3)}+(Bx+C) (3x+2)$

One way to move forward is to substitute $x=-2/3$, effectively zeroing the last term (which allows you to solve for $A$). Once we have $A$, unfortunately, we'll have to expand out all the terms and compare the coefficients of both sides of the equation.
 
  • #3
Let's use a better method;

(5x^2+1)/(3x+2)(x^2+3) = A/(3x+2) + (Bx+C)/(x^2+3)

To find A,multiplying both sides of the equality by (3x+2) ;

(5x^2+1)/(x^2+3) =A + (3x+2)(Bx+C)/(x^2+3) (1)

Now,setting x=-2/3 or 3x+2=0,makes the expression on the right containing (3x+2) vanish,so(5(-2/3)^2+1)/((-2/3)^2+3) =A + 0 .Then,

A=29/31To find C,set x=0 in (1),so we get rid of b.Then,

1/3=29/31 +2C/3

So,c can be easily calculated.

To find b,plugging any number but -2/3 & 0 into x,say 1,

6/5=29/31 + 5(B+C)/4

We have found C,then B can be easily calculated.
 
  • #4
Actually there are a number of mistakes- though probably mostly typos.

mathlearn said:
Trouble here in the below partial fraction (Bug)

$\frac{5x^2+1}{(3x+2)(x^2+3)}$

One factor in the denominator is a quadratic expression

Split this into two parts A&B

$\frac{5x^2+1}{(3x+2)(x^2+3)}=\frac{A}{(3x+2)}+\frac{Bx+c}{(x^2+3)}$
Okay.

$\frac{5x^2+1}{(3x+2)(x^2+3)}=\frac{A(x^2+3)}{(3x+2)}+\frac{Bx+c (3x+2)}{(x^2+3)}$
No, the fractions on the right must both have "$(x^2+ 3)(3x+ 2)$" as denominator and the final fraction should have numerator (Bx+c)(3x+ 2) not "Bx+ c(3x+ 2)".

\frac{5x^2+1}{(3x+2)(x^2+3)}=\frac{A(x^2+3)}{(3x+2)}+\frac{Bx+c (3x+2)}{(x^2+3)}$
${5x^2+1}={A(x^2+3)}+{Bx+c (3x+2)}{}$
Again that should be (Bx+ c)(3x+ 2).

and cannot take it from here
$5x^2+ 1= Ax^2+ 3A+ 3Bx^2+ 3cx+ 2Bx+ 2c$
$5x^2+ 0x+ 1= (A+ 3B)x^2+ (3c+ 2B)x+ (3A+ 2c)$
Since this is t0 be true for all x, "corresponding coefficients must be equal:
5= A+ 3B, 0= 3c+ 2B, and 1= 3A+ 2c

Subtract 3 times the second equation from twice the first:
10- 0= 2A+ 6B- 6c- 6B= 2A- 6c
Subtract 3(1= 3A+ 2c), 3= 9A- 6c from that:
7= -7A so A= -1. Then -3+ 2c= 1 so 2c= 4 and c= 2.
5= -1+ 3B so 3B= 6, B= 2.

A= -1, B= 2, and c= 2.
 

1. What are partial fractions?

Partial fractions is a method used to simplify a complex rational expression into smaller, simpler fractions. This is done by breaking down the denominator into smaller factors and expressing the original fraction as a sum of simpler fractions.

2. How do you solve for partial fractions?

To solve for partial fractions, you must first factor the denominator of the rational expression. Then, create a system of equations using the coefficients of each factor and solve for the unknown constants. Finally, substitute these constants back into the original expression to get the simplified partial fractions.

3. Why is partial fractions important?

Partial fractions is an important tool in mathematics and engineering, as it allows for the simplification of complex rational expressions. This can help in solving equations, integrating functions, and solving differential equations.

4. Can you use partial fractions to solve any rational expression?

No, not all rational expressions can be simplified using partial fractions. This method only works for proper rational expressions, which have a degree of the numerator that is less than the degree of the denominator. Improper rational expressions can be simplified using polynomial division.

5. Is there a shortcut or trick to solving partial fractions?

While there is no shortcut or trick to solving partial fractions, practice and familiarity with factoring and solving systems of equations can make the process easier and quicker. It is also important to check your work and make sure the simplified partial fractions are equivalent to the original rational expression.

Similar threads

Replies
1
Views
834
  • General Math
Replies
2
Views
766
Replies
3
Views
2K
Replies
19
Views
2K
  • General Math
Replies
2
Views
1K
Replies
6
Views
2K
  • General Math
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
881
  • Precalculus Mathematics Homework Help
Replies
1
Views
895
  • General Math
Replies
2
Views
1K
Back
Top